Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $11.3$ years; the standard deviation is $2.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living less than $15.7$ years.
$11.3$ $9.1$ $13.5$ $6.9$ $15.7$ $4.7$ $17.9$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $11.3$ years. We know the standard deviation is $2.2$ years, so one standard deviation below the mean is $9.1$ years and one standard deviation above the mean is $13.5$ years. Two standard deviations below the mean is $6.9$ years and two standard deviations above the mean is $15.7$ years. Three standard deviations below the mean is $4.7$ years and three standard deviations above the mean is $17.9$ years. We are interested in the probability of a lion living less than $15.7$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the lions will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the lions will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $6.9$ years and the other half $({2.5\%})$ will live longer than $15.7$ years. The probability of a particular lion living less than $15.7$ years is ${95\%} + {2.5\%}$, or $97.5\%$.